LintCode: Number of Islands

分析：经典连通分量问题

图：

　　节点：所有1的位置

　　边：两个相邻的1的位置有一条边

BFS／DFS (DFS使用递归，代码较短)

　　选一个没标记的点，然后搜索，扩展4个邻居（如果有），直到不能扩展

　　每一次是一个连通分量

　　难点：标记节点——判重

C++

DFS

class Solution {
public:
    void help(vector<vector<bool>>& a, int x, int y) {
        if ((x < 0) || (x >= a.size()) || (y < 0) || (y >= a[x].size()) || a[x][y] == false) {
            return ;
        }
        a[x][y] = false;
        help(a, x + 1, y);
        help(a, x - 1, y);
        help(a, x, y + 1);
        help(a, x, y - 1);
    }
    /**
     * @param grid a boolean 2D matrix
     * @return an integer
     */
    int numIslands(vector<vector<bool>>& grid) {
        // Write your code here
        int ans = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[i].size(); ++j) {
                if (grid[i][j] == true) {
                    help(grid, i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }
};

C++

BFS

class Solution {
public:
    void help(vector<vector<bool>>& a, int x, int y) {
        queue<pair<int, int> > q;
        const int dx[] = {-1, 1, 0, 0};
        const int dy[] = {0, 0, -1, 1};
        a[x][y] = false;
        for (q.push(make_pair(x, y)); !q.empty(); q.pop()) {
            x = q.front().first;
            y = q.front().second;
            for (int i = 0; i < 4; ++i) {
                int nx = x + dx[i];
                int ny = y + dy[i];
                if ((nx >= 0) && (nx < a.size()) && (ny >= 0) &&(ny < a[nx].size()) && (a[nx][ny] == true)) {
                    a[nx][ny] = false;
                    q.push(make_pair(nx, ny));
                }
            }
        }
    }
    /**
     * @param grid a boolean 2D matrix
     * @return an integer
     */
    int numIslands(vector<vector<bool>>& grid) {
        // Write your code here
        int ans = 0;
        for (int i = 0; i < grid.size(); ++i) {
            for (int j = 0; j < grid[i].size(); ++j) {
                if (grid[i][j] == true) {
                    help(grid, i, j);
                    ++ans;
                }
            }
        }
        return ans;
    }
};